AMU Engineering 2020 solved paper Solved Paper

When a spring is in equilibrium, its extension =‌

mg

k

At the highest point of block, spring is in its natural length.
Hence, amplitude, A=‌

mg

k

Given, f=10Hz
‌‌ω=2πf=20πrad∕s
∴‌‌ω=√‌

k

m

On squaring both sides, we get
‌

k

m

‌‌=ω2
⇒‌‌vmax‌‌=Aω
‌‌=‌

m

k

gω=‌

gω

ω2

∴‌‌A‌‌=‌

m

ω

=‌

π2

20π

=‌

π

20

m∕s
∴‌‌‌‌A‌‌=‌

mg

k

=‌

g

ω2

‌‌=‌

π2

400π2

=‌

0.25

100

m=0.25cm
Now, when extension is 0.20cm, then displacement from equilibrium position
‌‌=0.25−0.20=0.05cm
x‌‌=0.05cm
∴‌‌v‌‌=ω√A2−x2
‌‌=20π√(0.25)2−(0.05)2
‌‌=2π√6cm∕s
‌‌=15.4cm∕s